Question: If $8a + 6b + 4c = 4$ and $4x + 5y + 4z = 9$, what is $28c + 56a + 28z + 35y + 42b + 28x$ ?
Solution: $= 56a + 42b + 28c + 28x + 35y + 28z$ $= (7) \cdot (8a + 6b + 4c) + (7) \cdot (4x + 5y + 4z)$ $= (7) \cdot (4) + (7) \cdot (9)$ $= 28 + 63$ $= 91$